4

I have two questions:

1) I can only let the items visible when clicking the screen, just adding up <2->. However, I don't know how the let the contents in the items do the same thing. It's not good to let them appear at the same time.

2) To make the formula be left. See my code, I have tried two methods but both of them fail.

More details, please see the picture and my code.

enter image description here

\documentclass[CJK]{beamer} 
\usepackage{beamerthemesplit} 
\usepackage[english]{babel}
\setbeamercovered{transparent}
\usepackage{amsmath}
\begin{document}
\begin{frame}
\frametitle{16.Given$f(x)=\sqrt{2}\cos (x-\frac{\pi}{12})$,$x\in \mathbb{R}$.\\
1)$f(-\frac{\pi}{6})=?$;
2)If $\cos \theta=\frac{3}{5}$,$\theta \in(\frac{3\pi}{2},2\pi)$, $f(2\theta+\frac{\pi}{3})$=?.}
%\vspace{-1cm}%delete the gap
\begin{itemize} \parsep 10pt %\itemsep 10pt
\item<2-> $1)f(-\frac{\pi}{6})=\sqrt{2}\cos (-\frac{\pi}{6}-\frac{\pi}  {12})=\sqrt{2}\cos (-\frac{\pi}{4})=\sqrt{2}\times \frac{\sqrt{2}}{2}=1.$\\
\item<3-> 2) ∵$\cos \theta=\frac{3}{5},\frac{3\pi}{2}<\theta<2\pi$,
∴$\sin \theta=-\sqrt{1-cos^2\theta}=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$
∴ $\sin 2\theta=2\sin \theta \cos\theta=-\frac{24}{25}\Rightarrow \cos     2\theta=\cos^2\theta \sin^2\theta=-\frac{7}{25}$
 %\begin{flushleft} %fail to let the formula be left
 %f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})\\
 %&=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta\\
 %&=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.
 %\end{flushleft}
 \begin{flalign*} % fail again
 f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})\\
 &=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta\\
 &=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.
 \end{flalign*}
 \end{itemize}
 \end{frame}
 \end{document} 
  • 1
    For the first question: does this answer help? – Ludovic C. Nov 17 '13 at 12:39
4

Another method using the flalign environment would be to add && at the end of each line inside this environment. This combined with some \pause commands yields the following output:

Output

And the MWE:

\documentclass[CJK]{beamer} 

\usepackage{beamerthemesplit} 
\usepackage[english]{babel}
\setbeamercovered{transparent}
\usepackage{amsmath}

\setbeamercovered{transparent=0}

\begin{document}

\begin{frame}
\frametitle{16.Given $f(x)=\sqrt{2}\cos (x-\frac{\pi}{12})$,$x\in \mathbb{R}$.\\
1) $f(-\frac{\pi}{6})=?$;
2) If $\cos \theta=\frac{3}{5}$,$\theta \in(\frac{3\pi}{2},2\pi)$, $f(2\theta+\frac{\pi}{3})$=?}

\pause1) $f(-\frac{\pi}{6})=\sqrt{2}\cos (-\frac{\pi}{6}-\frac{\pi}  {12})=\sqrt{2}\cos (-\frac{\pi}{4})=\sqrt{2}\times \frac{\sqrt{2}}{2}=1.$\\
\pause2) $\cos \theta=\frac{3}{5},\frac{3\pi}{2}<\theta<2\pi$
\pause$\sin \theta=-\sqrt{1-cos^2\theta}=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$\\
\pause$\sin 2\theta=2\sin \theta \cos\theta=-\frac{24}{25}\Rightarrow \cos 2\theta=\cos^2\theta \sin^2\theta=-\frac{7}{25}$\pause
\begin{flalign}\nonumber
f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})&&\\\nonumber
&=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta&\\\nonumber
&=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.&&
\end{flalign}

\end{frame}
\end{document} 
  • Thank you for your reply! Your answer is quite good. Just one more thing. For the formula in the align, could we still use "\pause"? – Hinn Nov 17 '13 at 14:31
  • 1
    @Hinn Yes but you have to insert the \pause command between the & and the = sign (at least after the first & of each line). – Ludovic C. Nov 17 '13 at 14:34
3

regarding part 2 of your question, instead of the displayed flalign*, use ordinary in-line math (terminate the previous line with \\) and use aligned.

since aligned has a small extra space at the beginning (see Why is there a \, space at the beginning of the aligned environment?), begin the first line of the alignment with \! to negate it.

here's what that input would look like:

 \begin{itemize}
 ...
 \\
 $\begin{aligned}
  \! f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})\\
  &=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta\\
  &=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.
 \end{aligned}$
 \end{itemize}

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